And with that post, kylekatarn becomes a power member. Well done to you.

Back on topic, I tried it and got f(f(x)) = a (ax + b|x|) + b|(ax + b|x|)|.

For g(g(x)) to equal x when f(f(x)) is equal to x, then f(f(x) would have to be equal to g(g(x) and so b would have to be 0.

So now you have to prove that a (ax + b|x|) + b|(ax + b|x|)| ≠ x for all values of x when b ≠ 0. That's the tricky bit.

mathsyperson wrote:

And with that post, kylekatarn becomes a power member. Well done to you.

Btw, thanks! : )

...indeed:)
Again, thank you all:)

I finnaly got it. I don't know if you still want the solution but here it is, anyway:

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We want to show that if f( f(x) ) = x then g( g(x) ) = x, for all x∊ℝ

H1 (Hipothesis):
∀x∊ℝ f( f(x) ) = x

T1 (Thesis):
∀x∊ℝ g( g(x) ) = x


This is an implication where
H1 ⇒ T1
∀x∊ℝ [ f(f(x)) = x ⇒ g(g(x)) = x ]


We will determine the conditions for the Hipothesis to be true and then do the same for the Thesis. The we find the conditions are the same for both Hipothesis and Thesis. This allows us to conclude that the inicial statement is more than an implication, its something "logically stronger" - an equivalence H1⇔T1
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L1 (Lemma 1):
∀x≥0 ⇒ |x| = x
∀x<0 ⇒ |x| = -x
(the case x=0 was included in the first implication, but it could be on the second one, too)

Proof: Its obvious, considering the definition of absolute value.
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Proof of H1:
∀x∊ℝ f( f(x) ) = x

H1.a) Case x≥0
f(x) = ax+b|x|
f(x) = ax+bx     //by L1
f(x) = (a+b)x
f( f(x) ) = f( (a+b)x ) = (a+b)²x

f( f(x) ) = x
(a+b)²x = x ⇔ (a+b)² = 1

Conclusion from H1.a)
∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b)² = 1

H1.b) Case x<0
f(x) = ax+b|x|
f(x) = ax+b(-x)     //by L1
f(x) = ax-bx
f(x) = (a-b)x
f( f(x) ) = f( (a-b)x ) = (a-b)²x

f( f(x) ) = x
(a-b)²x = x ⇔ (a-b)² = 1

Conclusion from H1.b)
∀x<0 [ f( f(x) ) = x ] ⇔ (a-b)² = 1

Conclusion from H1.a) and H1.b)
∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b)² = 1
and
∀x<0 [ f( f(x) ) = x ] ⇔ (a-b)² = 1
therefore
∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b)² = (a-b)² =1 ]

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Proof of T1:
∀x∊ℝ g( g(x) ) = x

T1.a) Case x≥0
g(x) = gx-b|x| 
g(x) = ax-bx     //by L1
g(x) = (a-b)x
g(g(x)) = g((a-b)x) = (a-b)²x

g(g(x)) = x
(a-b)²x = x ⇔ (a-b)² = 1

Conclusion from T1.a)
∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b)² = 1

H1.b) Case x<0
g(x) = ax-b|x|
g(x) = ax-b(-x)     //by L1
g(x) = ax+bx
g(x) = (a+b)x
g( g(x) ) = f( (a+b)x ) = (a+b)²x

g( g(x) ) = x
(a+b)²x = x ⇔ (a+b)² = 1

Conclusion from T1.b)
∀x<0 [ g( g(x) ) = x ] ⇔ (a+b)² = 1

Conclusion from T1.a) and T1.b)
∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b)² = 1
and
∀x<0 [ g( g(x) ) = x ] ⇔ (a+b)² = 1
therefore
∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b)² = (a+b)² =1 ]

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Conclusions from H1 and T1
∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b)² = (a-b)² =1 ]
and
∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b)² = (a+b)² =1 ]
therefore
∀x∊ℝ [ f( f(x) ) = x ] ⇔ ∀x∊ℝ [ g( g(x) ) = x ]     //by the transitivity property of '⇔'
∀x∊ℝ [ f( f(x) ) = x ⇔ g( g(x) ) = x ]
so we can say that
H1⇔ T1 ∴ H1⇒ T1     //this is what we wanted to prove
and more..
T1⇒ H1

Last edited by kylekatarn (2005-11-06 03:46:17)

no problem : )